Question

5．可以將橢圓$\frac{{x}^{2}}{10}$+$\frac{{y}^{2}}{8}$=1變?yōu)閳Ax²+y²=4的伸縮變換為（　�。�

• A． $\left\{\begin{array}{l}{x′=\frac{2}{5}x}\\{y′=\frac{\sqrt{2}}{2}y}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x′=\frac{\sqrt{10}}{2}x}\\{y′=\sqrt{2}y}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x′=\frac{\sqrt{2}}{2}x}\\{y′=\frac{\sqrt{10}}{5}y}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x′=\frac{\sqrt{10}}{5}x}\\{y′=\frac{\sqrt{2}}{2}y}\end{array}\right.$

Answer 1

分析 令$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$代入，化簡代入橢圓方程化簡整理即可得出．

解答 解：由圓x²+y²=4化為$（\frac{x}{2}）^{2}+（\frac{y}{2}）^{2}$=1，
令$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$代入橢圓方程可得$\frac{（{x}^{′}）^{2}}{4}+\frac{（{y}^{′}）^{2}}{4}$=1，即（x′）²+（y′）²=4，
由$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$化為$\left\{\begin{array}{l}{{x}^{′}=\frac{\sqrt{10}}{5}x}\\{{y}^{′}=\frac{\sqrt{2}}{2}y}\end{array}\right.$．
故選：D．

點(diǎn)評(píng) 本題考查了橢圓化為圓的變換公式，考查了計(jì)算能力，屬于基礎(chǔ)題．